Algebraically closed group

In mathematics, in the realm of group theory, a group $A$ is algebraically closed if any finite set of equations and inequations that "make sense" in $A$ already have a solution in $A$. This idea will be made precise later in the article.

Informal discussion

Suppose we wished to find an element $x$ of a group $G$ satisfying the conditions (equations and inequations):

::$x^2=1$::$x^3=1$::$x\; e\; 1$

Then it is easy to see that this is impossible because the first two equations imply $x=1$. In this case we say the set of conditions are inconsistent with $G$. (In fact this set of conditions are inconsistent with any group whatsoever.)

However if we extend the group $G$ to the group $H$ with multiplication table:

Then the condions have two solutions, namely $x=b$ and $x=c$.

Thus there are three possibilities regarding such conditions:
* They may be inconsistent with $G$ and have no solution in any extension of $G$.
* They may have a solution in $G$.
* They may have no solution in $G$ but nevertheless have a solution in some extension $H$ of $G$.

It is reasonable to ask whether there are any groups $A$ such that whenever a set of conditions like these have a solution at all, they have a solution in $A$ itself? The answer turns out to be "yes", and we call such groups algebraically closed groups.

Formal definition of an algebraically closed group

We first need some preliminary ideas.

If $G$ is a group and $F$ is the free group on countably many generators, then by a finite set of equations and inequations with coefficients in $G$ we mean a pair of subsets $E$ and $I$ of $Fstar\; G$ the free product of $F$ and $G$.

This formalizes the notion of a set of equations and inequations consisting of variables $x\_i$ and elements $g\_j$ of $G$. The set $E$ represents equations like: ::$x\_1^2g\_1^4x\_3=1$::$x\_3^2g\_2x\_4g\_1=1$::$dots$The set $I$ represents inequations like::$g\_5^\{-1\}x\_3\; e\; 1$::$dots$

By a solution in $G$ to this finite set of equations and inequations, we mean a homomorphism $f:F\; ightarrow\; G$, such that $ilde\{f\}(e)=1$ for all $ein\; E$ and $ilde\{f\}(i)\; e\; 1$ for all $iin\; I$. Where $ilde\{f\}$ is the unique homomorphism $ilde\{f\}:Fstar\; G\; ightarrow\; G$ that equals $f$ on $F$ and is the identity on $G$.

This formalizes the idea of substituting elements of $G$ for the variables to get true identities and inidentities. In the example the substitutions $x\_1mapsto\; g\_6,\; x\_3mapsto\; g\_7$ and $x\_4mapsto\; g\_8$ yield:::$g\_6^2g\_1^4g\_7=1$::$g\_7^2g\_2g\_8g\_1=1$::$dots$::$g\_5^\{-1\}g\_7\; e\; 1$::$dots$

We say the finite set of equations and inequations is consistent with $G$ if we can solve them in a "bigger" group $H$. More formally:

The equations and inequations are consistent with $G$ if there is a group$H$ and an embedding $h:G\; ightarrow\; H$ such that the finite set of equations and inequations $ilde\{h\}(E)$ and $ilde\{h\}(I)$ has a solution in $H$. Where $ilde\{h\}$ is the unique homomorphism $ilde\{h\}:Fstar\; G\; ightarrow\; Fstar\; H$ that equals $h$ on $G$ and is the identity on $F$.

Now we formally define the group $A$ to be algebraically closed if every finite set of equations and inequations that has coefficients in $A$ and is consistent with $A$ has a solution in $A$.

Known Results

It is difficult to give concrete examples of algebraically closed groups as the following results indicate:

* Every countable group can be embedded in a countable algebraically closed group.
* Every algebraically closed group is simple.
* No algebraically closed group is finitely generated.
* An algebraically closed group cannot be recursively presented.
* A finitely generated group has solvable word problem if and only if it can embedded in every algebraically closed group.

The proofs of these results are, in general very complex. However a sketch the proof that a countable group $C$ can be embedded in an algebraically closed group follows.

First we embed $C$ in a countable group $C\_1$ with the property that every finite set of equations with coefficients in $C$ that is consistent in $C\_1$ has a solution in $C\_1$ as follows:

There are only countably many finite sets of equations and inequations with coefficients in $C$. Fix an enumeration $S\_0,S\_1,S\_2,dots$ of them. Define groups $D\_0,D\_1,D\_2,dots$ inductively by:

::$D\_0\; =\; C$

::

Now let:

::$C\_1=cup\_\{i=0\}^\{infty\}D\_\{i\}$

Now iterate this construction to get a sequence of groups $C=C\_0,C\_1,C\_2,dots$ and let:

::$A=cup\_\{i=0\}^\{infty\}C\_\{i\}$

Then $A$ is a countable group containing $C$. It is algebraically closed because any finite set of equations and inequations that is consistent with $A$ must have coefficients in some $C\_i$ and so must have a solution in $C\_\{i+1\}$.

References

* A. Macintyre: On algebraically closed groups, ann. of Math, 96, 53-97 (1972)
* B.H. Neumann: A note on algebraically closed groups. J. London Math. Soc. 27, 227-242 (1952)
* B.H. Neumann: The isomorphism problem for algebraically closed groups. In: Word Problems, pp 553-562. Amsterdam: North-Holland 1973
* W.R. Scott: Algebraically closed groups. Proc. Amer. Math. Soc. 2, 118-121 (1951)